0
0.45 repeated = 5/11
METHOD #1
0.45 repeated = 0.45 + 0.0045 + 0.000045 ...
r = 0.01 [common ratio between each term]
n = 0.45 [beginning term]
As |r|<1
0.45 + 0.0045 + 0.000045 ... forms a limiting sum
Limiting sum = S(n) = n / (1 - r)
= 0.45/(1 - 0.01)
= 0.45/0.99
= 45/99
= 5/11
METHOD #2
Let x = 0.45 repeated
100x = 45.45 repeated
100x - x = 45.454545454545... - 0.454545454545...
99x = 45
x = 45/99
x = 5/11
0.454545...
Let P = 0.454545....
Hence
100P = 45.454545...
Subtract
99P = 45. 0 = 45 ( NB THe recurring decimals to infinity subtract to zero.
99P = 45
P = 45/99
Cancel down by '9'
5/11
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